With the help of the following data,find out the change in heat content for the reaction:
$C_2H_{4(g)} + H_{2(g)} \to C_2H_{6(g)}$

Bond	Bond energy $(kJ \ mol^{-1})$
$C-H$	$413$
$C-C$	$348$
$C=C$	$610$
$H-H$	$436$

Assertion $(A)$: Formation of $H_2O_{(\ell)}$ from $H_{2(g)}$ and $O_{2(g)}$ is more exothermic than formation of $H_2O_{(g)}$ from $H_{2(g)}$ and $O_{2(g)}$.
Reason $(R)$: $H_2O_{(\ell)}$ has more attraction force than $H_2O_{(g)}$ and it is more stable than $H_2O_{(g)}$.

The heat of neutralization of $HCl$ by $NaOH$ under certain conditions is $-55.9 \, kJ \, mol^{-1}$ and that of $HCN$ by $NaOH$ is $-12.1 \, kJ \, mol^{-1}$. The heat of ionization of $HCN$ is .............. $kJ \, mol^{-1}$.

Given: $S + O_2 \rightarrow SO_2 : \Delta H_1 = -298.2 \ kJ$,$SO_2 + 1/2 \ O_2 \rightarrow SO_3 : \Delta H_2 = -98.7 \ kJ$,$SO_3 + H_2O \rightarrow H_2SO_4 : \Delta H_3 = -130.2 \ kJ$,and $H_2 + 1/2 \ O_2 \rightarrow H_2O : \Delta H_4 = -287.3 \ kJ$. Calculate the heat of formation of $H_2SO_4$ in $kJ$.

The molar neutralization heat for $KOH$ and $HNO_3$ as compared to the molar neutralization heat of $NaOH$ and $HCl$ is:

The standard enthalpies of formation for $CH_{4(g)}$,$CO_{2(g)}$,and $H_2O_{(l)}$ are $-75$,$-393.5$,and $-286 \ kJ \ mol^{-1}$ respectively. The enthalpy change for the reaction $CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}$ is ......... $kJ$.